The integral is this one: $\int{dy\over (y^2+b^2)^{3/2}}$. I know that the answer is $y\over b^2\sqrt{y^2+b^2}$ but I'd want to know how it's solve step by step. Thanks for your answers.
Calculate an indefinite-integral in function of a parameter
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indefinite-integrals
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0If the parameter troubles you, get rid of it by the change of variable $z=by$. – 2017-01-26
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0what is your problem exactly? do you find the solutions given below hard to follow? – 2017-01-26
3 Answers
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Let $y=b\tan(u)$ and $dy=b\sec^2(u)~du$. $$\int \frac{b\sec^2(u)}{(b^2\tan^2(u)+b^2)^{3/2}}~du$$ $$=\int \frac{b\sec^2(u)}{(b^2(\tan^2(u)+1))^{3/2}}~du$$ $$=\int \frac{b\sec^2(u)}{b^3\sec^3(u)}~du$$ We thus obtain: $$\frac{1}{b^2}\int \cos(u)~du$$ Solve this integral, and when substituting back from $y=b\tan(u)$, use the fact that $\sin(\arctan(x))\equiv \frac{x}{\sqrt{x^2+1}}$.
If done correctly, this should give the answer you've specified (plus the arbitrary constant of integration). Please let us know if you cannot get the answer from the above.
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Let us use the known integral
$$I_b:=\int\frac{dy}{\sqrt{y^2+b^2}}=\int\frac{dy}{b\sqrt{\dfrac{y^2}{b^2}+1}}=\text{arsinh}\frac yb+C_b$$
Then differentiating on $b$ under the integral,
$$I'_b=-\int\frac{b\,dy}{(y^2+b^2)^{3/2}}=-\frac y{b^2}\frac1{\sqrt{\dfrac{y^2}{b^2}+1}}+C'_b$$ and the claim follows.
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$$I=\int{dy\over (y^2+b^2)^{3/2}}$$ Let $$y=b \sinh(t)\implies dy=b\cosh(t)\,dt$$ This makes $$I=\frac{1}{b^2}\int\frac{dt}{\cosh^2(t)}=\frac{1}{b^2} \tanh (t)=\frac{1}{b^2} \frac{\sinh(t)}{\cosh (t)}=\frac{1}{b^2} \frac{\sinh(t)}{\sqrt{1+\sinh^2 (t)}}$$ Replace now $\sinh(t)$ by $\frac yb$ and simplify.